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Q.

Evaluate ∫222x22x2xdx

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a

=1(log⁡2)322x+c

b

=1(log⁡2)3222x+c

c

=1(log⁡2)32x+c

d

=1(log⁡2)2222x+c

answer is B.

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Detailed Solution

I=∫222x22x2xdx Let  222x=t or 222x22x2x(log⁡2)3dx=dt∴ I=∫1(log⁡2)3dt=1(log⁡2)3t+c=1(log⁡2)3222x+c
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